# Lewis Structures and Bonding (If we did it after molecular shape- AKA VSEPR- it would be a prequel to “What shape are your molecules in?”)

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Lewis Structures and Bonding (If we did it after molecular shape- AKA VSEPR- it would be a prequel to “What shape are your molecules in?”)

You’ll find out about…  Lewis structures How many bonds do each element make? What can expand? What can be deficient?  Bonding (covalent) Polarity  Electronegativity and determining bond type  Isomers  Formal charge  Resonance  Resonance v. isomers

Lewis Structures  Remember that all atoms want a full outer shell, so do Lewis structures  Remember that for a given Lewis structure, the number of electrons around the atoms must equal the total number of electrons individually assigned. Ex: C has 4, H has 1, so CH 4 must have 8 total We also did NH 4 + on the board

Drawing Lewis Structures Add the valence electrons. Identify the central atom (usually the one with the highest molecular mass and closest to the center of the periodic table). Place the central atom in the center of the molecule and add all other atoms around it. Place one bond (two electrons) between each pair of atoms. Complete the octet for the central atom. Complete the octets for all other atoms. Use double bonds if necessary.

Determining formal charge Formal charge can be determined by: Normal number of electrons in outer shell - [(1/2 the number of bonded electrons) + lone electrons] _____________________________________ = formal charge

Formal Charge, continued  Example: N in NH 4 +  FC =5- [(1/2 of 8)+ 0]= +1  H in NH 4 +  FC =1- [(1/2 of 2)+ 0]= 0  Overall, the formal charge on NH 4 + is +, so we write NH 4 + or as [NH 4 ] + This bracketed version is typically used, and is more precise for reasons we have yet to get into, but we will.

Formal charge and stability  The most “happy” molecules tend to have no formal charges  However, molecules may be “happy” if they have no NET charge on them if there is 1+ and 1-, so a net of +1 + (-1)=0  Structures that are the best have a minimal formal charge and a full octet (valence shell) around each atom

Electronegativity and Bond Type  Determine the absolute value between the electronegativities of the atoms involved If it is, ≥ 2 then the bond is ionic If it is 0.5 ≥1.9 then the bond is polar covalent  Electrons are shared unequally between the atoms resulting in partial charges (dipoles) If it is ≤ 0.4, then the bond is nonpolar covalent  Electrons are shared equally between the atoms of the bond no partial charges

Examples:  A bond from H-O H has an electronegativity of 2.1 O has an electronegativity of 3.5  The difference is 1.4, so the bond is polar The more electronegative element has a slightly negative charge (here, O) The less electronegative element has a slightly positive charge (here, H)

Ionic Bonding Na(s) + ½Cl 2 (g)  NaCl(s)  H° f = - 410.9 kJ NaCl is more stable than its constituent elements. Why? Na has lost an electron to become Na + and chlorine has gained the electron to become Cl . Na + (10 electrons) => 1s 2 2s 2 sp 6 same as Ne Cl - (18 electrons) => [Ne]3s 2 3p 6 same as Ar That is, both Na + and Cl  have noble gas configurations (an octet of electrons surrounding the central ion).

K-F Ionic ED = 3.2 Cl-F Covalent Polar ED = 1.0 F-F Covalent NonPolar ED = 0.0 Electronegativity and Bond Polarity

Covalent Bonding Multiple Bonds It is possible for more than one pair of electrons to be shared between two atoms (multiple bonds): One shared pair of electrons = single bond (e.g. H 2 ); Two shared pairs of electrons = double bond (e.g. O 2 ); Three shared pairs of electrons = triple bond (e.g. N 2 ). Generally, bond distances decrease as we move from single through double to triple bonds.

Bonds and Bond Order  A single bond has a bond order of 1  Multiple bonds double: two shared pairs of electrons  Bond order of a double bond is 2 triple: three shared pairs of electrons  Bond order of a triple bond is 3 quadruple bonds do not occur (too hard to share 8 electrons; steric hinderance)

Bond Energy  Bond energy: the amount of energy needed to break a bond Usually measured in KJ/ mol  Bond energy for each type of bond (Ex: C-C single, C=C double, C-O, C=O, H-H, C-N, C=N, etc.) is different Why? Different atomic radii for each element, different electronegativity, etc)

Bond Energy and Bond Order  Single bonds have lowest bond energy  Double bonds have higher bond energy  Triple bonds have highest bond energy

Bond energies of common bonds (kJ/ mol)  C-C376  C=C720  C C962  N N945

Atomic radii  H-F570  H-Cl432  H-Br366  Cl-Cl243  Br-Br193  I-I151

Bond Order and Bond Length BondBond OrderBond EnthalpyBond Length C-C1348 kJ/mol1.54 Å C=C2614 kJ/mol1.34 Å CCCC3839 kJ/mol1.20 Å N-N1163 kJ/mol1.47 Å N=N2418 kJ/mol1.24 Å NNNN3941 kJ/mol1.10 Å

Isomers  Same formula, different arrangement of atoms  Physically break bonds and MOVE atoms  Other example on the board; looked at structural v molecular formulas and the hybrid version with CH 3 OCH 3 and CH 2 OHCH 3

Resonance Structures  Have the same alignment of atoms, but different bonding (ONLY are electrons are moved, those both in bonds and lone pairs are switched)

Resonance structures of BF 3 Remember that none of these is a real picture of BF 3, but the real picture is a hybrid of all of these. (with BO=1.3 for all B-F bonds) Some of these are better pictures of what really happens than others: the better ones are those with the least formal charge.

AGAIN: Formal charge and stability  The most “happy*” molecules tend to have no formal charges  However, molecules may be “happy” if they have no NET charge on them if there is 1+ and 1-, so a net of +1 + (-1)=0  Resonance structures that are better structures have a minimal formal charge and a full octet (valence shell) around each atom * “happy” = stable and energetically favorable

Resonance, Formal Charge and Good Lewis Structures  Those that are not as good, but fulfill the octet rule are considered to be minor contributors They can exist, but not as much of the overall picture looks like them  If it doesn’t fulfill the octet rule and it isn’t a KNOWN exception*, and/ or it has crazy formal charges, toss it. It doesn’t work. (* exceptions: next slide)

Words About Exceptions:  They happen. A fair amount of the time with CERTAIN ELEMENTS- not with everything.  Hence, exceptions to the general rules, but rules for those elements are now different.

Words About Exceptions:  Q: Why do chemists do this to me? A: Because they hate you.  Q: Really? A: No, they don’t. The properties of the atoms set the rules, based upon their electron configurations and other properties discussed in class (IE, EA, AR, shielding, Z eff ….) Chemists do not make up rules to annoy you; they make up general rules to fit MOST situations; elements sometimes don’t fit the constraints.  Q: Will the exceptions mess me up? A: Not if you can follow all of the other rules.

Deficient* atoms: *(less than a full shell)  Boron : USUALLY keeps 6 electrons in its outer shell. It just does, because of being small and being a nonmetal.  Basically, every element below is a metal, and gives up electrons to bond ionically.  But B is so small it wants more. So it bonds covalently to get them. But it can’t go to a full 8- it’s too many.

Expansion of the octet  P, S, halogens and noble gases (yeah, I know- we’ll explain why later on) heavier than Br (Br, I, At, Kr, Xe, Rn)

Expand to fit…  Typically 5 or 6 PAIRS of electrons, and these are added as lone pairs, not as bonded pairs.  Add lone pairs on the central atom only until the number of electrons needed is reached.  Ex: PCl 5, I 3, SF 4, XeF 3

An excerpt from a web site on expansion:  The concept of the Expanded Octet occurs in any system that has an atom with more than four electron pairs attached to it.  Most commonly, atoms will expand their octets to contain a total of five or six electron pairs, in total. In theory, it is possible to expand beyond those number. The large amounts of negative charge concentrated in small volumes of space prevent those larger expanded octets from forming.  When an atom expands its octet, it does so by making use of empty d orbitals that are available in the valence level of the atom doing the expanding.  Atoms that do not have empty valence level d orbitals will not be able to expand their octets. The atom that expands its octet in a structure will usually be located in the center of the structure and the system will not use any multiple bonds in attaching atoms to the central atom.  The process of expanding octets is strictly a last resort on the part of atoms. http://www.bcpl.net/~kdrews/molegeo/molegeo.html#Expanded

More theft: But this is well stated at http://www.towson.edu/~yau/ExpandedOctet.htm http://www.towson.edu/~yau/ExpandedOctet.htm  “Expanded octet” refers to the Lewis structures where the central atom ends up with more than an octet, such as in PCl 5 or XeF 4. In drawing the Lewis structure for PCl5, there is a total of 40 valence electrons to put in (5 + 5x7 = 40). One can easily see that if the central atom, P, is to be joined to five Cl atoms, P would have 10 electrons instead of the octet. (Remember that one does NOT string out the elements to look thus: P-Cl-Cl-Cl-Cl-Cl.  No !!)  Clearly there is a violation of the “Octet Rule”.  How do we know to allow this violation? It’s simply this: PCl 5 exists. Our rules have to be revised to accommodate observations. If PCl 5 exists, then this violation must be permitted, since there is no other way to explain it. Again, chemists don’t hate you. Nature just does things that we didn’t plan on when we use the easiest explanations.

Exceptions continued…  A word of caution: Does this mean that we can violate the Octet Rule any time we wish? No! We can violate the Octet Rule only when there is no other way to explain it.  Limitations to the “expanded octet”:  This cannot occur with elements that do not contain d-orbitals! Which elements would that be? Elements smaller than neon (atomic number 10) have electrons only in the first two main energy levels (n = 1 and n = 2), and those energy levels do not contain d-orbitals.

More on exceptions…  Second of all, expanded octets generally occur when there are too many electrons to fit in. Since double bonds and triple bonds occur only when there are insufficient number of electrons, you would not normally apply the “expanded octet” to central atoms with multiple bonds! In other words, the extra electrons must be added as lone pairs and NOT as double or triple bonds!

Summary of limitations of applying the “expanded octet”:  The central atom with an expanded octet  MUST have an atomic number larger than 10 (beyond Ne) (this means it has available d orbitals to extend into)  Extra electrons should be first placed on the outside atoms to fulfill their octets. After that, there are still extra electrons, start with placing them as lone pairs on the central atom. If the central atom does not have a positive formal charge, do not go any further. You have the correct Lewis structure. Only if the central atom has a positive charge should you move a lone pair from the outside atoms to share (to neutralize the formal positive charge)  Do NOT indiscriminately double or triple bond!

Let’s make XeF 4  Step One: Count the total number of valence electrons. 8 from Xe+ 4 F x (7e- per F) = 36e -, or 18 pairs.  Step Two: The first element is usually the central atom, and then you cluster the other atoms around it.  Step Three: The 4 covalent bonds shown above account for 4 pairs. As you put lone pairs onto the surrounding F, you would account for 12 more pairs, giving you a total of 16 pairs. Where are you going to put two more pairs? The only place would be on the central atom (on Xe) as lone pairs. Xe would therefore have 4 atoms and a lone pair (AX 4 E 2 ).

 Xenon now has twelve electrons instead of the octet! This is called an “expanded octet”, expanding beyond the octet.  Notice we do NOT put a double bond between the Xe and one of the F! A mistake that students often make. In the structure shown above, Xe has a formal charge of zero, so there is no reason to do any more to the structure.

 This structure is incorrect because Xe now has a formal charge of 2+ (6 electrons instead of 8) and the two F with double bonds each has a formal charge of −1 (8 electrons instead of 7). Compared to the structure above which has no formal charges, this is certainly not preferred.  In addition, in terms of the Octet Rule, F has exceed the Octet and it is not an element than can exceed the Octet Rule (atomic number les than 10).

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